The frequency of revolution of an electron revolving in a circular path in a magnetic field of $7.8 × 10^{-4} T$ will be

(Take $e = -1.6 × 10^{-19} C; m_e = 9.1 × 10^{-31} kg; \pi = 22/7$)

21.8 MHz

The correct answer is Option (2) → 21.8 MHz

Given:

Magnetic field, $B = 7.8 \times 10^{-4}\ \text{T}$

Formula for frequency of revolution (cyclotron frequency):

$f = \frac{eB}{2\pi m_e}$

Substitute values:

$e = 1.6 \times 10^{-19}\ \text{C}$, $m_e = 9.1 \times 10^{-31}\ \text{kg}$

$f = \frac{1.6 \times 10^{-19} \times 7.8 \times 10^{-4}}{2\pi \times 9.1 \times 10^{-31}}$

$f = \frac{1.248 \times 10^{-22}}{5.72 \times 10^{-30}} = 2.18 \times 10^{7}\ \text{Hz}$

Answer: $f = 2.2 \times 10^{7}\ \text{Hz}$